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Question 3: Complementarity & Genetic Code.5 points A. 2 point. Imagine you have

ID: 194086 • Letter: Q

Question

Question 3: Complementarity & Genetic Code.5 points A. 2 point. Imagine you have a DNA double helix that is 200 bp in length (meaning both the top and bottom strands are each 200 nucleotides long). If this dsDNA molecule has 60 adenines altogether (counting the total number of As on the top and bottom strands), specify how many of each of the other nucleotides must exist in this molecule. If there is not enough information, specify "X # of thymines: # of cytosines: # of guanines; B. 1 points. As you know the genetic code is redundant' because most amino acids can be specified by more than one codon. For example, the amino acid proline can be specified by these 4 different codons: CCU, CCC, CCA, CCG. This means that there are 4 different tRNA genes in the genome that get transcribed and then conjugated to the amino acid proline. These 4 tRNAs each have a different anti-codon to match the codon sequences for proline List the 4 tRNA anti-codons that match codons for proline. Be sure to write each tRNA with the polarity as specified, 3' to 5 3' 5 5 3' 5 3' 5 C. 2 points. Imagine you decide to build an organism from scratch, using the same genetic code that we currently have. However, to save time and money, you decide to use only one of the possible codons for each amino acid. That means you only have to make 20 kinds of tRNA genes, one for each of the 20 amino acids. You will also have to reserve 1 codon to be the stop codon. Once you do this, how many unused codons will you have? Note: this question has real-world applications because you can use the unused codons to make an texpanded genetic code", e.g. see: https:/len wikipedia.org/wiki/ to derive your answer. Expanded genetic code" To get credit, show the equation you used

Explanation / Answer

Ans A: since total number of Adenine residues equals 60 , which will pair with 60 Thymine residues. So the number of G + C = 80. Since G and C pair with each other so guanine residues = 40 and cytosine residues = 40.

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